CS205b/CME306Lecture 21 Conservation of MassConservation of mass may be derived by examining the density of a control volume Ω with boun dary∂Ω as in Figure 1. The density at any point in space is ρ, and it moves with velocity u. The massof this control volume ismass =ZΩρ dV.Any change in this mass is due to material leaving and entering the control volume. Let n bethe outward-facing (unit) normal of a small patch of the boundary ∂Ω. Then, n · u is the rate ofmovement across the boundary at any point (with positive ind icating movement out of the volume).The remaining velocity component u − (n · u)n is due to flow along the boundary but not acrossit. If the the surface patch has area A, then the rate of m ass flow across the boundary patch isA(n · u)ρ = (ρu) · dS, where dS is the surface element. The total of this flow across the boundarythe the rate of mass decrease, since a positive value indicates flow out of the boundary. This leadsto the statement of mass conservation∂∂tZΩρ dV = −Z∂Ω(ρu) · dS.This is called the weak form, since it does not involve spatial derivatives. Strong form is obtainedby applying the divergence theorem to the right hand side.∂∂tZΩρ dV = −ZΩ∇ · (ρu) dV.Ω∂ΩFigure 1: Control volume Ω with boundary ∂Ω through which material flows.1Figure 2: Three chunks located at x1, x2, and x3. The black curves show the density contributionof a single chunk at each point in space. The d otted line sh ows the density profile of space.By moving the partial time derivative into the integral, the two integrals may be mergedZΩ∂ρ∂t+ ∇ · (ρu) dV = 0.Finally, this must be true of any control volume Ω, so that the strong form is obtainedρt+ ∇ · (ρu) = 0.If we restrict ourselves to 1D, we can derive this result in another way. In 1D, the controlvolume is an interval Ω = [a, b]. The rate of increase in material across the endpoint a is ρRuR,and th e rate of increase across the endpoint b is ρRuR, since positive velocity repr esents materialleaving the control volume. This expresses conservation of mass as∂∂tZΩρ dV =∂∂tZ[a,b]ρ dx = −(ρRuR− ρLuL).Defining interval width ∆x = b − a and the average densityρ =1∆xR[a,b]ρ dx, th is becomes∂ρ∂t= −ρRuR− ρLuL∆x.In the limit of a → b, we have ∆x → 0. If we let ρ = ρRand u = uR, thenρ → ρ. The limit of theright hand side is the classical definition of a derivative. Putting these together yields∂ρ∂t= −∂(ρu)∂x.2 Smoothed Particle Hydrodynamics (SPH)If we consider mass as attached to chunks, we automatically conserve mass as we move the chunksaround. However, because the laws governing m otion often involve derivatives of quantities storedon the chunks, it is useful to h ave a definition of these properties everywhere in space that issufficiently differentiable. This can be achieved by spreading attributes associated with chunks,such as their mass, over some local region of space. This is the basic idea behind the SPH method.Let W (x) be some sufficiently differentiable function that can be used to spread attributesover space. In particular, consider a chunk centered at xiwith mass mi. Then define the densitycontribution from this chunk at any point x in space to be ρ(x) = miW (x − xi). Because we wouldlike to conserve mass, we insist that th e total mass spread thr ough out space is the mass assignedto that chunkmi=Z∞−∞ρ(x) dx =Z∞−∞miW (x − xi) dxwhich leads to the requirementZ∞−∞W (x) dx = 1.It is also desirable for W (x) to be symmetric about the origin (so that density is spread evenly inall directions about its center of mass). It is also more efficient if W (x) has local influence in that2W (x) = 0 everywhere outside some region around the origin. Note that W (x) has units of one overvolume, since it yields unity when integrated over the volume of space.To be very u s eful, our m odel must contain many chunks. In this case, we define the density atany point as the sum of the density contributions of every chunk as illustrated in Figure 2.ρ(x) =XimiW (x −